-[[!toc]]
+[[!toc levels=2]]
-#Recursion: fixed points in the lambda calculus##
+**Chris:** I'll be working on this page heavily until 11--11:30 or so. Sorry not to do it last night, I crashed.
+
+
+#Recursion: fixed points in the Lambda Calculus#
Sometimes when you type in a web search, Google will suggest
alternatives. For instance, if you type in "Lingusitics", it will ask
you "Did you mean Linguistics?". But the engineers at Google have
added some playfulness to the system. For instance, if you search for
"anagram", Google asks you "Did you mean: nag a ram?" And if you
-search for "recursion", Google asks: "Did you mean: recursion?"
+[search for "recursion"](http://www.google.com/search?q=recursion), Google asks: "Did you mean: recursion?"
##What is the "rec" part of "letrec" doing?##
-How could we compute the length of a list? Without worrying yet about what lambda-calculus implementation we're using for the list, the basic idea would be to define this recursively:
+How could we compute the length of a list? Without worrying yet about what Lambda Calculus encoding we're using for the list, the basic idea is to define this recursively:
-> the empty list has length 0
+> the empty list has length 0
-> any non-empty list has length 1 + (the length of its tail)
+> any non-empty list has length 1 + (the length of its tail)
In OCaml, you'd define that like this:
- let rec get_length = fun lst ->
- if lst == [] then 0 else 1 + get_length (tail lst)
- in ... (* here you go on to use the function "get_length" *)
+ let rec length = fun xs ->
+ if xs == [] then 0 else 1 + length (tail xs)
+ in ... (* here you go on to use the function "length" *)
In Scheme you'd define it like this:
- (letrec [(get_length
- (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
- ... ; here you go on to use the function "get_length"
- )
+ (letrec [(length (lambda (xs)
+ (if (null? xs) 0
+ (+ 1 (length (cdr xs))) )))]
+ ... ; here you go on to use the function "length"
+ )
Some comments on this:
-1. `null?` is Scheme's way of saying `isempty`. That is, `(null? lst)` returns true (which Scheme writes as `#t`) iff `lst` is the empty list (which Scheme writes as `'()` or `(list)`).
-
-2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of an ordered pair. It just turns out to return the tail of a list because of the particular way Scheme implements lists.)
+1. `null?` is Scheme's way of saying `empty?`. That is, `(null? xs)` returns true (which Scheme writes as `#t`) iff `xs` is the empty list (which Scheme writes as `'()` or `(list)`).
-3. I use `get_length` instead of the convention we've been following so far of hyphenated names, as in `make-list`, because we're discussing OCaml code here, too, and OCaml doesn't permit the hyphenated variable names. OCaml requires variables to always start with a lower-case letter (or `_`), and then continue with only letters, numbers, `_` or `'`. Most other programming languages are similar. Scheme is very relaxed, and permits you to use `-`, `?`, `/`, and all sorts of other crazy characters in your variable names.
+2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of a [[dotted pair|week3_unit#imp]]. As we discussed in notes for last week, it just turns out to return the tail of a list because of the particular way Scheme implements lists.)
-4. I alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
+3. We alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
The main question for us to dwell on here is: What are the `let rec` in the OCaml code and the `letrec` in the Scheme code?
-Answer: These work like the `let` expressions we've already seen, except that they let you use the variable `get_length` *inside* the body of the function being bound to it---with the understanding that it will there refer to the same function that you're then in the process of binding to `get_length`. So our recursively-defined function works the way we'd expect it to. In OCaml:
+Answer: These work a lot like `let` expressions, except that they let you use the variable `length` *inside* the body of the function being bound to it---with the understanding that it will there be bound to *the same function* that you're *then* in the process of binding `length` to. So our recursively-defined function works the way we'd expect it to. Here is OCaml:
- let rec get_length = fun lst ->
- if lst == [] then 0 else 1 + get_length (tail lst)
- in get_length [20; 30]
- (* this evaluates to 2 *)
+ let rec length = fun xs ->
+ if xs == [] then 0 else 1 + length (tail xs)
+ in length [20; 30]
+ (* this evaluates to 2 *)
-In Scheme:
+Here is Scheme:
+
+ (letrec [(length (lambda (xs)
+ (if (null? xs) 0
+ (+ 1 (length (cdr xs))) )))]
+ (length (list 20 30)))
+ ; this evaluates to 2
- (letrec [(get_length
- (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
- (get_length (list 20 30)))
- ; this evaluates to 2
-
If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:
- let get_length = fun lst ->
- if lst == [] then 0 else 1 + get_length (tail lst)
- in get_length [20; 30]
- (* fails with error "Unbound value length" *)
+ let length = fun xs ->
+ if xs == [] then 0 else 1 + length (tail xs)
+ in length [20; 30]
+ (* fails with error "Unbound value length" *)
Here's Scheme:
- (let* [(get_length
- (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
- (get_length (list 20 30)))
- ; fails with error "reference to undefined identifier: get_length"
+ (let* [(length (lambda (xs)
+ (if (null? xs) 0
+ (+ 1 (length (cdr xs))) )))]
+ (length (list 20 30)))
+ ; fails with error "reference to undefined identifier: length"
Why? Because we said that constructions of this form:
- let get_length = A
- in B
+ let
+ length match/= A
+ in B
really were just another way of saying:
- (\get_length. B) A
+ (\length. B) A
-and so the occurrences of `get_length` in A *aren't bound by the `\get_length` that wraps B*. Those occurrences are free.
+and so the occurrences of `length` in A *aren't bound by the `\length` that wraps B*. Those occurrences are free.
-We can verify this by wrapping the whole expression in a more outer binding of `get_length` to some other function, say the constant function from any list to the integer 99:
+We can verify this by wrapping the whole expression in a more outer binding of `length` to some other function, say the constant function from any list to the integer `99`:
- let get_length = fun lst -> 99
- in let get_length = fun lst ->
- if lst == [] then 0 else 1 + get_length (tail lst)
- in get_length [20; 30]
- (* evaluates to 1 + 99 *)
+ let length = fun xs -> 99
+ in let length = fun xs ->
+ if xs == [] then 0 else 1 + length (tail xs)
+ in length [20; 30]
+ (* evaluates to 1 + 99 *)
-Here the use of `get_length` in `1 + get_length (tail lst)` can clearly be seen to be bound by the outermost `let`.
+Here the use of `length` in `1 + length (tail xs)` can clearly be seen to be bound by the outermost `let`.
-And indeed, if you tried to define `get_length` in the lambda calculus, how would you do it?
+And indeed, if you tried to define `length` in the Lambda Calculus, how would you do it?
- \lst. (isempty lst) zero (add one (get_length (extract-tail lst)))
+ \xs. (empty? xs) 0 (succ (length (tail xs)))
-We've defined all of `isempty`, `zero`, `add`, `one`, and `extract-tail` in earlier discussion. But what about `get_length`? That's not yet defined! In fact, that's the very formula we're trying here to specify.
+We've defined all of `empty?`, `0`, `succ`, and `tail` in earlier discussion. But what about `length`? That's not yet defined! In fact, that's the very formula we're trying here to specify.
What we really want to do is something like this:
- \lst. (isempty lst) zero (add one (... (extract-tail lst)))
+ \xs. (empty? xs) 0 (succ (... (tail xs)))
where this very same formula occupies the `...` position:
- \lst. (isempty lst) zero (add one (
- \lst. (isempty lst) zero (add one (... (extract-tail lst)))
- (extract-tail lst)))
+ \xs. (empty? xs) 0 (succ (\xs. (empty? xs) 0 (succ (... (tail xs)))
+ (tail xs)))
but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
So how could we do it? And how do OCaml and Scheme manage to do it, with their `let rec` and `letrec`?
-1. OCaml and Scheme do it using a trick. Well, not a trick. Actually an impressive, conceptually deep technique, which we haven't yet developed. Since we want to build up all the techniques we're using by hand, then, we shouldn't permit ourselves to rely on `let rec` or `letrec` until we thoroughly understand what's going on under the hood.
+1. OCaml and Scheme do it using a trick. Well, not a trick. Actually an impressive, conceptually deep technique, which we haven't yet developed. Since we want to build up all the techniques we're using by hand, then, we shouldn't permit ourselves to rely on `let rec` or `letrec` until we thoroughly understand what's going on under the hood.
-2. If you tried this in Scheme:
+2. If you tried this in Scheme:
- (define get_length
- (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )
-
- (get_length (list 20 30))
+ (define length (lambda (xs)
+ (if (null? xs) 0
+ (+ 1 (length (cdr xs))) )))
+
+ (length (list 20 30))
- You'd find that it works! This is because `define` in Scheme is really shorthand for `letrec`, not for plain `let` or `let*`. So we should regard this as cheating, too.
+ You'd find that it works! This is because `define` in Scheme is really shorthand for `letrec`, not for plain `let` or `let*`. So we should regard this as cheating, too.
-3. In fact, it *is* possible to define the `get_length` function in the lambda calculus despite these obstacles. This depends on using the "version 3" implementation of lists, and exploiting its internal structure: that it takes a function and a base value and returns the result of folding that function over the list, with that base value. So we could use this as a definition of `get_length`:
+3. In fact, it *is* possible to define the `length` function in the Lambda Calculus despite these obstacles. This depends on using the "version 3" encoding of lists, and exploiting its internal structure: that it takes a function and a base value and returns the result of folding that function over the list, with that base value. So we could use this as a definition of `length`:
- \lst. lst (\x sofar. successor sofar) zero
+ \xs. xs (\x sofar. successor sofar) 0
- What's happening here? We start with the value zero, then we apply the function `\x sofar. successor sofar` to the two arguments <code>x<sub>n</sub></code> and `zero`, where <code>x<sub>n</sub></code> is the last element of the list. This gives us `successor zero`, or `one`. That's the value we've accumuluted "so far." Then we go apply the function `\x sofar. successor sofar` to the two arguments <code>x<sub>n-1</sub></code> and the value `one` that we've accumulated "so far." This gives us `two`. We continue until we get to the start of the list. The value we've then built up "so far" will be the length of the list.
+ What's happening here? We start with the value `0`, then we apply the function `\x sofar. successor sofar` to the two arguments <code>x<sub>n</sub></code> and `0`, where <code>x<sub>n</sub></code> is the last element of the list. This gives us `successor 0`, or `1`. That's the value we've accumuluted "so far." Then we go apply the function `\x sofar. successor sofar` to the two arguments <code>x<sub>n-1</sub></code> and the value `1` that we've accumulated "so far." This gives us `two`. We continue until we get to the start of the list. The value we've then built up "so far" will be the length of the list.
We can use similar techniques to define many recursive operations on
lists and numbers. The reason we can do this is that our "version 3,"
-fold-based implementation of lists, and Church's implementations of
+fold-based encoding of lists, and Church's encodings of
numbers, have a internal structure that *mirrors* the common recursive
operations we'd use lists and numbers for. In a sense, the recursive
-structure of the `get_length` operation is built into the data
+structure of the `length` operation is built into the data
structure we are using to represent the list. The non-recursive
-version of get_length exploits this embedding of the recursion into
+version of length exploits this embedding of the recursion into
the data type.
This is one of the themes of the course: using data structures to
encode the state of some recursive operation. See discussions of the
[[zipper]] technique, and [[defunctionalization]].
-As we said before, it does take some ingenuity to define functions like `extract-tail` or `predecessor` for these implementations. However it can be done. (And it's not *that* difficult.) Given those functions, we can go on to define other functions like numeric equality, subtraction, and so on, just by exploiting the structure already present in our implementations of lists and numbers.
+As we said before, it does take some ingenuity to define functions like `tail` or `predecessor` for these encodings. However it can be done. (And it's not *that* difficult.) Given those functions, we can go on to define other functions like numeric equality, subtraction, and so on, just by exploiting the structure already present in our encodings of lists and numbers.
With sufficient ingenuity, a great many functions can be defined in the same way. For example, the factorial function is straightforward. The function which returns the nth term in the Fibonacci series is a bit more difficult, but also achievable.
isn't well-defined. The point of interest here is that its definition
requires recursion in the function definition.)
-Neither do the resources we've so far developed suffice to define the
+Neither do the resources we've so far developed suffice to define the
[[!wikipedia Ackermann function]]:
- A(m,n) =
- | when m == 0 -> n + 1
- | else when n == 0 -> A(m-1,1)
- | else -> A(m-1, A(m,n-1))
+ A(m,n) =
+ | when m == 0 -> n + 1
+ | else when n == 0 -> A(m-1,1)
+ | else -> A(m-1, A(m,n-1))
- A(0,y) = y+1
- A(1,y) = 2+(y+3) - 3
- A(2,y) = 2(y+3) - 3
- A(3,y) = 2^(y+3) - 3
- A(4,y) = 2^(2^(2^...2)) [where there are y+3 2s] - 3
- ...
+ A(0,y) = y+1
+ A(1,y) = 2+(y+3) - 3
+ A(2,y) = 2(y+3) - 3
+ A(3,y) = 2^(y+3) - 3
+ A(4,y) = 2^(2^(2^...2)) [where there are y+3 2s] - 3
+ ...
Many simpler functions always *could* be defined using the resources we've so far developed, although those definitions won't always be very efficient or easily intelligible.
###Fixed points###
-In general, we call a **fixed point** of a function f any value *x*
-such that f <em>x</em> is equivalent to *x*. For example,
-consider the squaring function `sqare` that maps natural numbers to their squares.
+In general, a **fixed point** of a function `f` is any value `x`
+such that `f x` is equivalent to `x`. For example,
+consider the squaring function `square` that maps natural numbers to their squares.
`square 2 = 4`, so `2` is not a fixed point. But `square 1 = 1`, so `1` is a
-fixed point of the squaring function.
+fixed point of the squaring function.
There are many beautiful theorems guaranteeing the existence of a
fixed point for various classes of interesting functions. For
instance, imainge that you are looking at a map of Manhattan, and you
are standing somewhere in Manhattan. The the [[!wikipedia Brouwer
-fixed point]] guarantees that there is a spot on the map that is
+fixed-point theorem]] guarantees that there is a spot on the map that is
directly above the corresponding spot in Manhattan. It's the spot
where the blue you-are-here dot should be.
point. (See the discussion below concerning a way of understanding
the successor function on which it does have a fixed point.)
-In the lambda calculus, we say a fixed point of an expression `f` is any formula `X` such that:
+In the Lambda Calculus, we say a fixed point of a term `f` is any term `X` such that:
- X <~~> f X
+ X <~~> f X
You should be able to immediately provide a fixed point of the
-identity combinator I. In fact, you should be able to provide a whole
-bunch of distinct fixed points.
+identity combinator I. In fact, you should be able to provide a
+whole bunch of distinct fixed points.
With a little thought, you should be able to provide a fixed point of
the false combinator, KI. Here's how to find it: recall that KI
What about K? Does it have a fixed point? You might not think so,
after trying on paper for a while.
-However, it's a theorem of the lambda calculus that every formula has
+However, it's a theorem of the Lambda Calculus that every formula has
a fixed point. In fact, it will have infinitely many, non-equivalent
fixed points. And we don't just know that they exist: for any given
formula, we can explicit define many of them.
-Yes, even the formula that you're using the define the successor
-function will have a fixed point. Isn't that weird? Think about how it
-might be true. We'll return to this point below.
+Yes, as we've mentioned, even the formula that you're using the define
+the successor function will have a fixed point. Isn't that weird?
+Think about how it might be true. We'll return to this point below.
-###How fixed points help definie recursive functions###
+###How fixed points help define recursive functions###
-Recall our initial, abortive attempt above to define the `get_length` function in the lambda calculus. We said "What we really want to do is something like this:
+Recall our initial, abortive attempt above to define the `length` function in the Lambda Calculus. We said "What we really want to do is something like this:
- \list. if empty list then zero else add one (... (tail lst))
+ \xs. if empty? xs then 0 else succ (... (tail xs))
where this very same formula occupies the `...` position."
Imagine replacing the `...` with some function that computes the
length function. Call that function `length`. Then we have
- \list. if empty list then zero else add one (length (tail lst))
+ \xs. if empty? xs then 0 else succ (length (tail xs))
At this point, we have a definition of the length function, though
it's not complete, since we don't know what value to use for the
symbol `length`. Technically, it has the status of an unbound
-variable.
+variable.
-Imagine now binding the mysterious variable:
+Imagine now binding the mysterious variable, and calling the resulting
+function `h`:
- h := \length \list . if empty list then zero else add one (length (tail list))
+ h := \length \xs. if empty? xs then 0 else succ (length (tail xs))
Now we have no unbound variables, and we have complete non-recursive
definitions of each of the other symbols.
-Let's call this function `h`. Then `h` takes an argument, and returns
-a function that accurately computes the length of a list---as long as
-the argument we supply is already the length function we are trying to
-define. (Dehydrated water: to reconstitute, just add water!)
+So `h` takes an argument, and returns a function that accurately
+computes the length of a list---as long as the argument we supply is
+already the length function we are trying to define. (Dehydrated
+water: to reconstitute, just add water!)
-But this is just another way of saying that we are looking for a fixed point.
-Assume that `h` has a fixed point, call it `LEN`. To say that `LEN`
-is a fixed point means that
+Here is where the discussion of fixed points becomes relevant. Saying
+that `h` is looking for an argument (call it `LEN`) that has the same
+behavior as the result of applying `h` to `LEN` is just another way of
+saying that we are looking for a fixed point for `h`.
h LEN <~~> LEN
-But this means that
+Replacing `h` with its definition, we have
+
+ (\xs. if empty? xs then 0 else succ (LEN (tail xs))) <~~> LEN
- (\list . if empty list then zero else add one (LEN (tail list))) <~~> LEN
+If we can find a value for `LEN` that satisfies this constraint, we'll
+have a function we can use to compute the length of an arbitrary list.
+All we have to do is find a fixed point for `h`.
-So at this point, we are going to search for fixed point.
The strategy we will present will turn out to be a general way of
finding a fixed point for any lambda term.
-##Deriving Y, a fixed point combinator##
+##Deriving Y, a fixed point combinator##
How shall we begin? Well, we need to find an argument to supply to
`h`. The argument has to be a function that computes the length of a
length of a list. Let's try applying `h` to itself. It won't quite
work, but examining the way in which it fails will lead to a solution.
- h h <~~> \list . if empty list then zero else 1 + h (tail list)
+ h h <~~> \xs. if empty? xs then 0 else 1 + h (tail xs)
-There's a problem. The diagnosis is that in the subexpression `h
-(tail list)`, we've applied `h` to a list, but `h` expects as its
-first argument the length function.
+The problem is that in the subexpression `h (tail list)`, we've
+applied `h` to a list, but `h` expects as its first argument the
+length function.
So let's adjust h, calling the adjusted function H:
- H = \h \list . if empty list then zero else one plus ((h h) (tail list))
+ H = \h \xs. if empty? xs then 0 else 1 + ((h h) (tail xs))
-This is the key creative step. Since `h` is expecting a
-length-computing function as its first argument, the adjustment
-tries supplying the closest candidate avaiable, namely, `h` itself.
+This is the key creative step. Instead of applying `h` to a list, we
+apply it first to itself. After applying `h` to an argument, it's
+ready to apply to a list, so we've solved the problem just noted.
+We're not done yet, of course; we don't yet know what argument to give
+to `H` that will behave in the desired way.
-We now reason about `H`. What exactly is H expecting as its first
-argument? Based on the excerpt `(h h) (tail l)`, it appears that `H`'s
-argument, `h`, should be a function that is ready to take itself as an
-argument, and that returns a function that takes a list as an
+So let's reason about `H`. What exactly is H expecting as its first
+argument? Based on the excerpt `(h h) (tail l)`, it appears that
+`H`'s argument, `h`, should be a function that is ready to take itself
+as an argument, and that returns a function that takes a list as an
argument. `H` itself fits the bill:
- H H <~~> (\h \list . if empty list then zero else 1 + ((h h) (tail list))) H
- <~~> \list . if empty list then zero else 1 + ((H H) (tail list))
- == \list . if empty list then zero else 1 + ((\list . if empty list then zero else 1 + ((H H) (tail list))) (tail list))
- <~~> \list . if empty list then zero
- else 1 + (if empty (tail list) then zero else 1 + ((H H) (tail (tail list))))
-
+ H H <~~> (\h \xs. if empty? xs then 0 else 1 + ((h h) (tail xs))) H
+ <~~> \xs. if empty? xs then 0 else 1 + ((H H) (tail xs))
+ == \xs. if empty? xs then 0 else 1 + ((\xs. if empty? xs then 0 else 1 + ((H H) (tail xs))) (tail xs))
+ <~~> \xs. if empty? xs then 0
+ else 1 + (if empty? (tail xs) then 0 else 1 + ((H H) (tail (tail xs))))
+
We're in business!
-How does the recursion work?
+How does the recursion work?
We've defined `H` in such a way that `H H` turns out to be the length function.
In order to evaluate `H H`, we substitute `H` into the body of the
lambda term. Inside the lambda term, once the substitution has
Since `H H` turns out to be the length function, we can think of `H`
by itself as half of the length function (which is why we called it
-`H`, of course). Given the implementation of addition as function
-application for Church numerals, this (H H) is quite literally H + H.
-Can you think up a recursion strategy that involves "dividing" the
-recursive function into equal thirds `T`, such that the length
-function <~~> T T T?
+`H`, of course). Can you think up a recursion strategy that involves
+"dividing" the recursive function into equal thirds `T`, such that the
+length function <~~> T T T?
We've starting with a particular recursive definition, and arrived at
a fixed point for that definition.
Works!
+Let's do one more example to illustrate. We'll do `K`, since we
+wondered above whether it had a fixed point.
+
+Before we begin, we can reason a bit about what the fixed point must
+be like. We're looking for a fixed point for `K`, i.e., `\xy.x`. `K`
+ignores its second argument. That means that no matter what we give
+`K` as its first argument, the result will ignore the next argument
+(that is, `KX` ignores its first argument, no matter what `X` is). So
+if `KX <~~> X`, `X` had also better ignore its first argument. But we
+also have `KX == (\xy.x)X ~~> \y.X`. This means that if `X` ignores
+its first argument, then `\y.X` will ignore its first two arguments.
+So once again, if `KX <~~> X`, `X` also had better ignore at least its
+first two arguments. Repeating this reasoning, we realize that `X`
+must be a function that ignores an infinite series of arguments.
+Our expectation, then, is that our recipe for finding fixed points
+will build us a function that somehow manages to ignore an infinite
+series of arguments.
+
+ h := \xy.x
+ H := \f.h(ff) == \f.(\xy.x)(ff) ~~> \fy.ff
+ H H := (\fy.ff)(\fy.ff) ~~> \y.(\fy.ff)(\fy.ff)
+
+Let's check that it is in fact a fixed point:
+
+ K(H H) == (\xy.x)((\fy.ff)(\fy.ff)
+ ~~> \y.(\fy.ff)(\fy.ff)
+
+Yep, `H H` and `K(H H)` both reduce to the same term.
+
+To see what this fixed point does, let's reduce it a bit more:
+
+ H H == (\fy.ff)(\fy.ff)
+ ~~> \y.(\fy.ff)(\fy.ff)
+ ~~> \yy.(\fy.ff)(\fy.ff)
+ ~~> \yyy.(\fy.ff)(\fy.ff)
+
+Sure enough, this fixed point ignores an endless, infinite series of
+arguments. It's a write-only memory, a black hole.
+
+Now that we have one fixed point, we can find others, for instance,
+
+ (\fy.fff)(\fy.fff)
+ ~~> \y.(\fy.fff)(\fy.fff)(\fy.fff)
+ ~~> \yy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)
+ ~~> \yyy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)
+
+Continuing in this way, you can now find an infinite number of fixed
+points, all of which have the crucial property of ignoring an infinite
+series of arguments.
+
+##What is a fixed point for the successor function?##
+
+As we've seen, the recipe just given for finding a fixed point worked
+great for our `h`, which we wrote as a definition for the length
+function. But the recipe doesn't make any assumptions about the
+internal structure of the function it works with. That means it can
+find a fixed point for literally any function whatsoever.
-##What is a fixed point for the successor function?##
+In particular, what could the fixed point for the
+successor function possibly be like?
Well, you might think, only some of the formulas that we might give to the `successor` as arguments would really represent numbers. If we said something like:
- successor make-pair
+ successor make-pair
who knows what we'd get back? Perhaps there's some non-number-representing formula such that when we feed it to `successor` as an argument, we get the same formula back.
Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the successor function.
-Moreover, the recipes that enable us to name fixed points for any
-given formula aren't *guaranteed* to give us *terminating* fixed
-points. They might give us formulas X such that neither `X` nor `f X`
-have normal forms. (Indeed, what they give us for the square function
-isn't any of the Church numerals, but is rather an expression with no
-normal form.) However, if we take care we can ensure that we *do* get
-terminating fixed points. And this gives us a principled, fully
-general strategy for doing recursion. It lets us define even functions
-like the Ackermann function, which were until now out of our reach. It
-would also let us define arithmetic and list functions on the "version
-1" and "version 2" implementations, where it wasn't always clear how
-to force the computation to "keep going."
+One (by now obvious) upshot is that the recipes that enable us to name
+fixed points for any given formula aren't *guaranteed* to give us
+*terminating* fixed points. They might give us formulas X such that
+neither `X` nor `f X` have normal forms. (Indeed, what they give us
+for the square function isn't any of the Church numerals, but is
+rather an expression with no normal form.) However, if we take care we
+can ensure that we *do* get terminating fixed points. And this gives
+us a principled, fully general strategy for doing recursion. It lets
+us define even functions like the Ackermann function, which were until
+now out of our reach. It would also let us define arithmetic and list
+functions on the "version 1" and "version 2" encodings, where it
+wasn't always clear how to force the computation to "keep going."
+
+###Varieties of fixed-point combinators###
OK, so how do we make use of this?
When you try to evaluate the application of that to some argument `M`, it's going to try to give you back:
- (\n. self n) M
+ (\n. self n) M
where `self` is equivalent to the very formula `\n. self n` that contains it. So the evaluation will proceed:
- (\n. self n) M ~~>
- self M ~~>
- (\n. self n) M ~~>
- self M ~~>
- ...
+ (\n. self n) M ~~>
+ self M ~~>
+ (\n. self n) M ~~>
+ self M ~~>
+ ...
You've written an infinite loop!
However, when we evaluate the application of our:
-<pre><code>Ψ (\self (\lst. (isempty lst) zero (add one (self (extract-tail lst))) ))</code></pre>
+<pre><code>Ψ (\self (\xs. (empty? xs) 0 (succ (self (tail xs))) ))</code></pre>
to some list `L`, we're not going to go into an infinite evaluation loop of that sort. At each cycle, we're going to be evaluating the application of:
- \lst. (isempty lst) zero (add one (self (extract-tail lst)))
+ \xs. (empty? xs) 0 (succ (self (tail xs)))
-to *the tail* of the list we were evaluating its application to at the previous stage. Assuming our lists are finite (and the implementations we're using don't permit otherwise), at some point one will get a list whose tail is empty, and then the evaluation of that formula to that tail will return `zero`. So the recursion eventually bottoms out in a base value.
+to *the tail* of the list we were evaluating its application to at the previous stage. Assuming our lists are finite (and the encodings we're using don't permit otherwise), at some point one will get a list whose tail is empty, and then the evaluation of that formula to that tail will return `0`. So the recursion eventually bottoms out in a base value.
##Fixed-point Combinators Are a Bit Intoxicating##
As we said, there are many other fixed-point combinators as well. For example, Jan Willem Klop pointed out that if we define `L` to be:
- \a b c d e f g h i j k l m n o p q s t u v w x y z r. (r (t h i s i s a f i x e d p o i n t c o m b i n a t o r))
+ \a b c d e f g h i j k l m n o p q s t u v w x y z r. (r (t h i s i s a f i x e d p o i n t c o m b i n a t o r))
then this is a fixed-point combinator:
- L L L L L L L L L L L L L L L L L L L L L L L L L L
+ L L L L L L L L L L L L L L L L L L L L L L L L L L
##Watching Y in action##
sink true true false ~~> I
sink true true true false ~~> I
-So we make `sink = Y (\f b. b f I)`:
+So we make `sink = Y (\f b. b f I)`:
- 1. sink false
+ 1. sink false
2. Y (\fb.bfI) false
3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) false
4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) false
Now we try the next most complex example:
- 1. sink true false
+ 1. sink true false
2. Y (\fb.bfI) true false
3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) true false
4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) true false
You should be able to see that `sink` will consume as many `true`s as
we throw at it, then turn into the identity function after it
-encounters the first `false`.
+encounters the first `false`.
The key to the recursion is that, thanks to Y, the definition of
`sink` contains within it the ability to fully regenerate itself as
That's why it's crucial to declare that 0! = 1, in which case the
recursive rule does not apply. In our terms,
- fac = Y (\fac n. iszero n 1 (fac (predecessor n)))
+ fac = Y (\fac n. zero? n 1 (fac (predecessor n)))
If `n` is 0, `fac` reduces to 1, without computing the recursive case.
Without pretending to give a serious analysis of the paradox, let's
assume that sentences can have for their meaning boolean functions
like the ones we have been working with here. Then the sentence *John
-is John* might denote the function `\x y. x`, our `true`.
+is John* might denote the function `\x y. x`, our `true`.
Then (1) denotes a function from whatever the referent of *this
sentence* is to a boolean. So (1) denotes `\f. f true false`, where
Y C
(\f. (\h. f (h h)) (\h. f (h h))) I
- (\h. C (h h)) (\h. C (h h)))
+ (\h. C (h h)) (\h. C (h h)))
C ((\h. C (h h)) (\h. C (h h)))
C (C ((\h. C (h h))(\h. C (h h))))
C (C (C ((\h. C (h h))(\h. C (h h)))))
So the Y combinator is only guaranteed to give us one fixed point out
of infinitely many---and not always the intuitively most useful
-one. (For instance, the squaring function has zero as a fixed point,
-since 0 * 0 = 0, and 1 as a fixed point, since 1 * 1 = 1, but `Y
-(\x. mul x x)` doesn't give us 0 or 1.) So with respect to the
+one. (For instance, the squaring function has `0` as a fixed point,
+since `0 * 0 = 0`, and `1` as a fixed point, since `1 * 1 = 1`, but `Y
+(\x. mul x x)` doesn't give us `0` or `1`.) So with respect to the
truth-teller paradox, why in the reasoning we've
just gone through should we be reaching for just this fixed point at
just this juncture?
One obstacle to thinking this through is the fact that a sentence
normally has only two truth values. We might consider instead a noun
-phrase such as
+phrase such as
(3) the entity that this noun phrase refers to
The chameleon nature of (3), by the way (a description that is equally
good at describing any object), makes it particularly well suited as a
-gloss on pronouns such as *it*. In the system of
+gloss on pronouns such as *it*. In the system of
[Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
pronouns denote (you guessed it!) identity functions...